An $$\alpha$$ particle and a carbon 12 atom has same kinetic energy K. The ratio of their de-Broglie wavelengths $$({\lambda _\alpha }:{\lambda _{C12}})$$ is :
A metal surface is illuminated by a radiation of wavelength 4500 $$\mathop A\limits^o $$. The ejected photo-electron enters a constant magnetic field of 2 mT making an angle of 90$$^\circ$$ with the magnetic field. If it starts revolving in a circular path of radius 2 mm, the work function of the metal is approximately :
An electron with speed v and a photon with speed c have the same de-Broglie wavelength. If the kinetic energy and momentum of electron are Ee and pe and that of photon are Eph and pph respectively. Which of the following is correct?
A proton, a neutron, an electron and an $$\alpha$$ particle have same energy. If $$\lambda$$p, $$\lambda$$n, $$\lambda$$e and $$\lambda$$a are the de Broglie's wavelengths of proton, neutron, electron and $$\alpha$$ particle respectively, then choose the correct relation from the following :